b,
$NH_3+H_2O\rightleftharpoons NH_4^++OH^-$
$C_{NH_3\text{phân li}}=0,05.1,88\%=9,4.10^{-4}M=[NH_4^+]=[OH^-]$
$\to [NH_3]=0,05-9,4.10^{-4}=0,04906M$
c,
$HF\rightleftharpoons H^++F^-$
Đặt $[H^+]=[F^-]=x(M)$
$C_{HF}=1(M)\to [HF]=1-x(M)$
$K_a=\dfrac{[H^+][F^-]}{[HF]}$
$\to x^2=6,3.10^{-4}(1-x)$
$\to x=0,0248$
Vậy: $[H^+]=[F^-]=0,0248M; [HF]=0,9752M$